show the inequality (a raised to(c-a))*(b raised to(a-b))*(c raised to (b-c))=or0 a=or

If use logarithms, you should be able to do the first
step. Use of logarithms helps you to reverse the
exponentiation.

log(a^(c-a)*b^(a-b)*c^(b-c))=<log
1

Use property of logarithms and transform the product in a
sum of logarithms.

log(a^(c-a))+ log(b^(a-b))+
log(c^(b-c))=<0 (use log1=0)

Use one of the
four logarithmic properties: the log of the power is equal to the log of the base
multiplied by power.

(c-a)log a + (a-b)log b+ (b-c) log
c=<0

Use a<b<c => log a<
log b< log c( because logarithmic function is monotonically
increasing)

Use c-a>=a-b and add a both sides
=> c>=2a-b.

Use
2a=<b+c

Use  Cebishev's inequality to (c-a)log a +
(a-b)log b=< (c-b)(log a)/2+ (c-b)(log b)/2 = (c-b)(log
ab)/2

Use the logarithmic property: the log of the power is
equal to the log of the base multiplied by power:(c-b)(log ab)/2 = (c-b)(log
sqrt(ab))

(c-a)log a + (a-b)log b+ (b-c) log
c=<(c-b)(log sqrt(ab))+ (b-c) log c=< 0

The
right side hand is (c-b)(log sqrt(ab))+ (b-c) log c=(c-b)(log sqrt(ab))- (c-b) log c=
(c-b)log (sqrt(ab)/c) =< 0

Check each factor of
product (c-b)log (sqrt(ab)/c) is is positive or negative: b<c =>
0<c-b =>c-b
positive

sqrt(ab)<c=>log (sqrt(ab)/c)
=> (c-b)log (sqrt(ab)/c) =< 0

Answer:
Inequality (c-b)log (sqrt(ab)/c) =< 0
proves (a^(c-a)*b^(a-b)*c^(b-c))=<1.